4
$\begingroup$

We have the definition:

Definition. Let $X$ be a scheme. Let $Z,Y⊂X$ be closed subschemes corresponding to quasi-coherent ideal sheaves $\mathcal{I},\mathcal{J}⊂\mathcal{O}_X$. The scheme theoretic intersection of $Z$ and $Y$ is the closed subscheme of $X$ cut out by $\mathcal{I}+\mathcal{J}$. The scheme theoretic union of Z and Y is the closed subscheme of $X$ cut out by $\mathcal{I}∩\mathcal{J}$.

Is true, in general, that given closed subschemes $Y,Z$ and $W$ of $X$ we have

$$Y\cap (Z\cup W)= (Y\cap Z)\cup (Y\cap W), \mbox{scheme-theoretically?}$$

If not, there is any necessary and sufficient conditions? Any hint, reference or solution is welcome!!

Remark: Using the language of ideal, if we denote $I_*$ the ideal of the variety $*$ we have just one inclusion, in general $$I_Y+(I_Z\cap I_W)\subseteq (I_Y+I_Z)\cap(I_Y+I_W).$$

$\endgroup$
3
  • 1
    $\begingroup$ As you say, the corresponding statement about ideals is not true in general. A counterexample there translates to a counterexample in affine schemes. $\endgroup$
    – Zhen Lin
    Aug 4, 2020 at 22:25
  • $\begingroup$ Exactly @ZhenLin, but any hint about a necessary and sufficient condition for this be true? $\endgroup$
    – IMP
    Aug 4, 2020 at 22:55
  • $\begingroup$ For example, again, talking about ideals. If the deals $I_Y+I_Z$ and $I_Y+I_W$ are co-prime, that is, if they sum up to the unitary ideal. Then the statement is true. In other words it is a sufficient condition. $\endgroup$
    – IMP
    Aug 4, 2020 at 23:18

1 Answer 1

0
$\begingroup$

No this is not ture, Professor Vakil's FOAG 2022version exercise9.1 I provides an example:

taking $V(y^2-x^2),V(y)\subset \Bbb{A}^2$ we see that $V(x^2-y^2) = V(x-y)\cup V(x+y)$. If we take $X= V(x-y), Y = V(x+y), Z = V(y)$ we see that $$(X\cup Y)\cap Z =\text{Spec }k[x,y]/(x^2-y^2,y) = \text{Spec }k[x,y]/(x^2 ,y) \cong \text{Spec } k[x]/(x^2) \tag{1}$$

While $$(X\cap Z)\cup (Y\cap Z) = \text{Spec } k[x,y]/(x,y) \cong \text{Spec } k \tag{2}$$

We see as a set they are the same, but as scheme, they are not the same, the fuzz information may lose in (2)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .